Simultaneous linear & quadratic equations

You should compare the coefficients in the two equations, and look for like terms: if you collect those terms, which terms will cancel?

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is elimination. The elimination method works by cancelling like terms which the two equations share. We need to figure out which of the terms in the equations will cancel, and why that cancellation happens.

In these two questions, there are two $x$-terms and two $y$-terms:

The goal of elimination is to cancel out like terms by adding (or subtracting) the equations. That can be either the $x$-terms or the $y$-terms.

For the equations in this question, notice that the coefficients of $y$ are equal and opposite. So if we add them together, they will cancel. So we will add the equations together to make that cancellation happen, like this:

The $y$-terms cancel each other out - that is what 'elimination' means! We get the $0y$ because the $y$-coefficients are equal and opposite: $-3y + \left(3y\right)$ is equal to zero. So the elimination depends on which of the coefficients cancel out.

The correct answer choices are: We will eliminate the $y$-terms because the $y$-coefficients will cancel.

You should compare the coefficients in the two equations, and look for like terms: if you collect those terms, which terms will cancel?

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is elimination. The elimination method works by cancelling like terms which the two equations share. We need to figure out which of the terms in the equations will cancel, and why that cancellation happens.

In these two questions, there are two $x$-terms and two $y$-terms:

The goal of elimination is to cancel out like terms by adding (or subtracting) the equations. That can be either the $x$-terms or the $y$-terms.

For the equations in this question, notice that the coefficients of $x$ are equal and opposite. So if we add them together, they will cancel. So we will add the equations together to make that cancellation happen, like this:

The $x$-terms cancel each other out - that is what 'elimination' means! We get the $0x$ because the $x$-coefficients are equal and opposite: $-3x + \left(3x\right)$ is equal to zero. So the elimination depends on which of the coefficients cancel out.

The correct answer choices are: We will eliminate the $x$-terms because the $x$-coefficients will cancel.

You should compare the coefficients in the two equations, and look for like terms: if you collect those terms, which terms will cancel?

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is elimination. The elimination method works by cancelling like terms which the two equations share. We need to figure out which of the terms in the equations will cancel, and why that cancellation happens.

In these two questions, there are two $x$-terms and two $y$-terms:

The goal of elimination is to cancel out like terms by adding (or subtracting) the equations. That can be either the $x$-terms or the $y$-terms.

For the equations in this question, notice that the coefficients of $x$ are equal and opposite. So if we add them together, they will cancel. So we will add the equations together to make that cancellation happen, like this:

The $x$-terms cancel each other out - that is what 'elimination' means! We get the $0x$ because the $x$-coefficients are equal and opposite: $3x + \left(-3x\right)$ is equal to zero. So the elimination depends on which of the coefficients cancel out.

The correct answer choices are: We will eliminate the $x$-terms because the $x$-coefficients will cancel.

The first thing to do is compare the coefficients of the terms. We can see that the coefficient of the $m$-term in the first equation is a multiple of the second $m$-term: the $\text{6}$ is $\text{2}$ times as much as $\text{12}$. If we multiply the entire second equation by this factor of $\text{2}$ we will be able to cancel the $m$-terms between the equations:

Now eliminate the $m$-terms. Since the terms have the same signs, we must subtract one equation from the other to cancel those terms.

Now we can solve for $n$:

Finally, use the value of $n$ to find the value of $m$. Remember that we can use either of the equations to do this, we can pick the easier equation. The equations are pretty much the same - for example, neither has any fractions or negatives. So we will use the first one.

The answers to the pair of equations are $m = -1$ and $n = 6$.

The first thing to do is compare the coefficients of the terms. We can see that the coefficient of the $f$-term in the first equation is a multiple of the second $f$-term: the $\text{3}$ is $\text{4}$ times as much as $\text{12}$. If we multiply the entire second equation by this factor of $\text{4}$ we will be able to cancel the $f$-terms between the equations:

Now eliminate the $f$-terms. Since the terms have opposite signs, we must add one equation to the other to cancel those terms.

Now we can solve for $g$:

Finally, use the value of $g$ to find the value of $f$. Remember that we can use either of the equations to do this, we can pick the easier equation. The second equation looks like a better choice because it has fewer negative signs than the first equation (in fact it does not have any!).

The answers to the pair of equations are $f = -2$ and $g = 2$.

The first thing to do is compare the coefficients of the terms. We can see that the coefficient of the $k$-term in the first equation is a multiple of the second $k$-term: the $\text{4}$ is $\text{2}$ times as much as $\text{8}$. If we multiply the entire second equation by this factor of $\text{2}$ we will be able to cancel the $k$-terms between the equations:

Now eliminate the $k$-terms. Since the terms have the same signs, we must subtract one equation from the other to cancel those terms.

Now we can solve for $j$:

Finally, use the value of $j$ to find the value of $k$. Remember that we can use either of the equations to do this, we can pick the easier equation. The second equation looks like a better choice because it has fewer negative signs than the first equation.

The answers to the pair of equations are $j = 7$ and $k = 4$.

Notice that the $y$-coefficients are opposites. You should start by adding the equations together, which will cancel those terms out.

We have two equations and we need to solve them 'simultaneously'. That means we need to find two numbers, for $x$ and for $y$, which solve both of the equations.

There are two methods for solving equations simultaneously: elimination and substitution. The equations in this question are perfect for elimination because the $y$-coefficients are equal and opposite. If we add the equations together, the $y$-terms will cancel. (We could solve these equations using the substitution method but elimination will be faster.)

So we add the equations together. This means we add together the left sides of the equations and also the right sides of the equations.

Now we have an equation with only one variable, and we can solve it.

So $x = -3$. But we are not done yet: we also need an answer for $y$. We can get this using the answer we just got for $x$. Substitute this into either of the equations to get the answer - we can use either equation because we are looking for the number that solves both of them. We will pick the second equation, because it has fewer negative signs.

Now we have the complete answer: the numbers which solve the equations are $x = -3$ and $y = -5$. We can see this represented beautifully on a graph of the two equations: the answers we just found are the coordinates of the point where the lines intersect.

The values which solve these two equations are $x = -3$ and $y = -5$.

Notice that the $y$-coefficients are opposites. You should start by adding the equations together, which will cancel those terms out.

We have two equations and we need to solve them 'simultaneously'. That means we need to find two numbers, for $x$ and for $y$, which solve both of the equations.

There are two methods for solving equations simultaneously: elimination and substitution. The equations in this question are perfect for elimination because the $y$-coefficients are equal and opposite. If we add the equations together, the $y$-terms will cancel. (We could solve these equations using the substitution method but elimination will be faster.)

So we add the equations together. This means we add together the left sides of the equations and also the right sides of the equations.

Now we have an equation with only one variable, and we can solve it.

So $x = -2$. But we are not done yet: we also need an answer for $y$. We can get this using the answer we just got for $x$. Substitute this into either of the equations to get the answer - we can use either equation because we are looking for the number that solves both of them. We will pick the second equation, because it has no negative signs.

Now we have the complete answer: the numbers which solve the equations are $x = -2$ and $y = 3$. We can see this represented beautifully on a graph of the two equations: the answers we just found are the coordinates of the point where the lines intersect.

The values which solve these two equations are $x = -2$ and $y = 3$.

Notice that the $y$-coefficients are opposites. You should start by adding the equations together, which will cancel those terms out.

We have two equations and we need to solve them 'simultaneously'. That means we need to find two numbers, for $x$ and for $y$, which solve both of the equations.

There are two methods for solving equations simultaneously: elimination and substitution. The equations in this question are perfect for elimination because the $y$-coefficients are equal and opposite. If we add the equations together, the $y$-terms will cancel. (We could solve these equations using the substitution method but elimination will be faster.)

So we add the equations together. This means we add together the left sides of the equations and also the right sides of the equations.

Now we have an equation with only one variable, and we can solve it.

So $x = -3$. But we are not done yet: we also need an answer for $y$. We can get this using the answer we just got for $x$. Substitute this into either of the equations to get the answer - we can use either equation because we are looking for the number that solves both of them. We will pick the first equation, because it has fewer negative signs.

Now we have the complete answer: the numbers which solve the equations are $x = -3$ and $y = 1$. We can see this represented beautifully on a graph of the two equations: the answers we just found are the coordinates of the point where the lines intersect.

The values which solve these two equations are $x = -3$ and $y = 1$.

You can solve the equations simultaneously using elimination or substitution. The first thing you should do is decide which method is the better choice for these equations.

To solve the given equations simultaneously, we can use either substitution or elimination. The best choice depends on the arrangement of the equations. These equations invite elimination, because the $y$-terms have equal coefficients. We can eliminate the $y$-terms by subtracting the equations.

Eliminate the $y$ and solve for $x$:

Super - we know that $x = -9$.

Now we can use the $x$-value to find the value of $y$. We can do this using either of the equations in the question. If one of the equations is simpler or more convenient (for example, if it has an isolated variable or has fewer negative signs) it is better to choose that equation. In this case the equations are pretty similar, so we will just pick the first one.

The answer is the pair of numbers $x = -9$ and $y = -5$. On the Cartesian plane, this looks like this:

The values which solve these two equations are $x = -9$ and $y = -5$.

You can solve the equations simultaneously using elimination or substitution. The first thing you should do is decide which method is the better choice for these equations.

To solve the given equations simultaneously, we can use either substitution or elimination. The best choice depends on the arrangement of the equations. These equations are not arranged nicely for elimination or substitution. And it is not possible to make a simple change to these equations. We need to multiply both of the equations so that we can get terms which will cancel. Based on the coefficients, one choice is to multiply the first equation by $\text{3}$ and the second equation by $\text{2}$. Then we can cancel the $x$-terms.

First change the first and second equation to set up the elimination:

and

Now do the elimination and complete the solution to find $y$.

Super - we know that $y = 1$.

Now we can use the $y$-value to find the value of $x$. We can do this using either of the equations in the question. If one of the equations is simpler or more convenient (for example, if it has an isolated variable or has fewer negative signs) it is better to choose that equation. In this case, the second equation is better because it has fewer negative signs.

The answer is the pair of numbers $x = -5$ and $y = 1$. On the Cartesian plane, this looks like this:

The values which solve these two equations are $x = -5$ and $y = 1$.

You can solve the equations simultaneously using elimination or substitution. The first thing you should do is decide which method is the better choice for these equations.

To solve the given equations simultaneously, we can use either substitution or elimination. The best choice depends on the arrangement of the equations. In this case we need to modify the equations before we can do elimination or substitution. And we only need to make a small change to arrange the equations for substitution: we just need to subtract the $-2*y$ term to the other side of the second equation.

First rearrange the second equation to set up the substitution:

Now do the substitution and complete the solution to find $y$.

Super - we know that $y = 0$.

Now we can use the $y$-value to find the value of $x$. We can do this using either of the equations in the question. If one of the equations is simpler or more convenient (for example, if it has an isolated variable or has fewer negative signs) it is better to choose that equation. In this case, the second equation is better because it has fewer negative signs.

The answer is the pair of numbers $x = 5$ and $y = 0$. On the Cartesian plane, this looks like this:

The values which solve these two equations are $x = 5$ and $y = 0$.

You need to solve these equations using the substitution method. So the first thing to do is pick one of the variables to isolate.

The question tells us to solve the equations using the substitution method. That is not necessarily the best method, but we do not have a choice!

None of the variables are isolated, so we will need to isolate one of them for the substitution step. It would be convenient if there was a coefficient of $\text{1}$ or $-\text{1}$, because that would make our work easier. But that is not the case. So we will just make $x$ the subject of the first equation:

Unfortunately, this equation has $\text{1}$ fraction in it. That is ugly, but it will work.

Substitute the expression for $x$ into the second equation. Then simplify and solve for the value of $y$:

Substitute this value of $y$ back into one of the equations and solve for $x$. In this case, the first equation has only $\text{2}$ negative signs while the other equation has $\text{3}$ negatives. So we will use the first equation.

The final answers are $x = \dfrac{67}{13}$ and $y = - \dfrac{90}{13}$.

You need to solve these equations using the substitution method. So the first thing to do is pick one of the variables to isolate.

The question tells us to solve the equations using the substitution method. That is not necessarily the best method, but we do not have a choice!

None of the variables are isolated, so we will need to isolate one of them for the substitution step. It would be convenient if there was a coefficient of $\text{1}$ or $-\text{1}$, because that would make our work easier. But that is not the case. So we will just make $x$ the subject of the first equation:

Unfortunately, this equation has $\text{1}$ fraction in it. That is ugly, but it will work.

Substitute the expression for $x$ into the second equation. Then simplify and solve for the value of $y$:

Substitute this value of $y$ back into one of the equations and solve for $x$. In this case, the equations are similar: they both have $\text{1}$ negative sign. We will pick the first equation.

The final answers are $x = - \dfrac{7}{4}$ and $y = - \dfrac{13}{6}$.

You need to solve these equations using the substitution method. So the first thing to do is pick one of the variables to isolate.

The question tells us to solve the equations using the substitution method. That is not necessarily the best method, but we do not have a choice!

None of the variables are isolated, so we will need to isolate one of them for the substitution step. It would be convenient if there was a coefficient of $\text{1}$ or $-\text{1}$, because that would make our work easier. But that is not the case. So we will just make $x$ the subject of the first equation:

Unfortunately, this equation has $\text{2}$ fractions in it. That is ugly, but it will work.

Substitute the expression for $x$ into the second equation. Then simplify and solve for the value of $y$:

Substitute this value of $y$ back into one of the equations and solve for $x$. In this case, we should use the first equation because it has no negative signs.

The final answers are $x = - \dfrac{5}{12}$ and $y = \dfrac{5}{4}$.

The first thing you should do is rearrange the first equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is $jk$.

We need to solve the equations simultaneously. To do that, we should start by rearranging the first equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by $jk$, which is the LCD of the fractions. That will cancel all of the denominators.

Now we are ready to solve the following equations:

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the first equation by $\text{2}$ so that the $k$-coefficients will be ready to cancel. Then we can eliminate them by subtracting the equations.

First modify things to set up the elimination:

Now do the elimination and complete the solution to find $j$.

Great - we have the first value, $j = 10$.

Now the last step: find the value of $k$. We can do this using either of the equations in the question: but the second eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the first equation).

The answer is the pair of numbers $j = 10$ and $k = 5$. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

The values which solve these two equations are $j = 10$ and $k = 5$.

The first thing you should do is rearrange the second equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is $xy$.

We need to solve the equations simultaneously. To do that, we should start by rearranging the second equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by $xy$, which is the LCD of the fractions. That will cancel all of the denominators.

Now we are ready to solve the following equations:

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations invite elimination, because the $y$-terms have equal and opposite coefficients. We can eliminate the $y$-terms by adding the equations.

Eliminate the $y$ and solve for $x$:

Great - we have the first value, $x = -2$.

Now the last step: find the value of $y$. We can do this using either of the equations in the question: but the first eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the second equation).

The answer is the pair of numbers $x = -2$ and $y = - \frac{5}{3}$. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

The values which solve these two equations are $x = -2$ and $y = - \frac{5}{3}$.

The first thing you should do is rearrange the second equation to remove the fractions. To do this, you should multiply the equation by the lowest common denominator of the fractions, which is $ab$.

We need to solve the equations simultaneously. To do that, we should start by rearranging the second equation. We definitely want to get rid of those fractions! We can do that if we multiply the expression by $ab$, which is the LCD of the fractions. That will cancel all of the denominators.

Now we are ready to solve the following equations:

We can use either elimination or substitution. It is important to decide which method is the easiest choice. Remember that the best choice is usually based on how the equations compare to each other.

These equations are not arranged nicely for elimination or substitution. We must change the equations somehow. In this case, we can make a small change that will allow us to use elimination: we can multiply the first equation by $\text{3}$ so that the $b$-coefficients will be ready to cancel. Then we can eliminate them by adding the equations.

First modify things to set up the elimination:

Now do the elimination and complete the solution to find $a$.

Great - we have the first value, $a = -3$.

Now the last step: find the value of $b$. We can do this using either of the equations in the question: but the first eqaution is probably easier because it does not include fractions (and because each variable only occurs once, not twice, like in the second equation).

The answer is the pair of numbers $a = -3$ and $b = 9$. As always, the answers we just found are the coordinates of the point where the lines intersect on the Cartesian plane.

The values which solve these two equations are $a = -3$ and $b = 9$.

We need to solve the equations in this equation using elimination. (Substitution will also work, but the question says that we should use elimination.) As always for elimination, we need to use the two equations to cancel some of the terms.

By comparing the coefficients of the terms, we can see that the second $y$-term has a coefficient which is a multiple of the first $y$-term. In fact, $\text{3}$ is $\text{3}$ times as much as $\text{1}$. We can make both coefficients of $y$ the same by using this factor of $\text{3}$ to multiply the entire first equation:

Now we can eliminate the $y$-terms by subtracting one equation from the other. Then we can solve for $x$.

Now we can solve for $x$:

The last step is to substitute $y$ back into either of the equations so that we can find $x$. Here we will use the first equation:

The answers are $x = 7$ and $y = -4$.

We need to solve the equations in this equation using elimination. (Substitution will also work, but the question says that we should use elimination.) As always for elimination, we need to use the two equations to cancel some of the terms.

By comparing the coefficients of the terms, we can see that the second $x$-term has a coefficient which is a multiple of the first $x$-term. In fact, $\text{6}$ is $\text{6}$ times as much as $\text{1}$. We can make both coefficients of $x$ the same by using this factor of $\text{6}$ to multiply the entire first equation:

Now we can eliminate the $x$-terms by subtracting one equation from the other. Then we can solve for $y$.

Now we can solve for $y$:

The last step is to substitute $y$ back into either of the equations so that we can find $x$. Here we will use the first equation:

The answers are $x = -8$ and $y = 7$.

We need to solve the equations in this equation using elimination. (Substitution will also work, but the question says that we should use elimination.) As always for elimination, we need to use the two equations to cancel some of the terms.

By comparing the coefficients of the terms, we can see that the second $y$-term has a coefficient which is a multiple of the first $y$-term. In fact, $\text{10}$ is $\text{2}$ times as much as $\text{5}$. We can make both coefficients of $y$ the same by using this factor of $\text{2}$ to multiply the entire first equation:

Now we can eliminate the $y$-terms by subtracting one equation from the other. Then we can solve for $x$.

Now we can solve for $x$:

The last step is to substitute $y$ back into either of the equations so that we can find $x$. Here we will use the first equation:

The answers are $x = -3$ and $y = -2$.

The best choice for substitution is based on which variable is easiest to isolate. In this case, you should look for the variable which is already isolated.

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is substitution. The substitution method works by using one of the equations to replace one of the variables in the other equation. We need to figure out which variable is the best choice to substitute for these two equations.

In these two equations, there are two $x$-terms and two $y$-terms:

The goal of substitution is to use one of the equations to remove a variable from the other equation.

In this case, the best choice is to substitute the $y$ from the first equation into the second equation. This is because in the first equation the $y$ is already isolated. To use substitution, we need a variable which is isolated (alone on one side of the equation).

In this case, the substitution looks like this:

Combining these equations by substitution, we get:

Since the first equation tells us that $y$ is equal to $y=- x + 5$, we can substitute it into the other equation in place of $y$ straight away.

The best choice is to substitute the $y$ from the first equation into the second equation because it is already isolated.

The best choice for substitution is based on which variable is easiest to isolate. In this case, you should look for the variable which is already isolated.

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is substitution. The substitution method works by using one of the equations to replace one of the variables in the other equation. We need to figure out which variable is the best choice to substitute for these two equations.

In these two equations, there are two $x$-terms and two $y$-terms:

The goal of substitution is to use one of the equations to remove a variable from the other equation.

In this case, the best choice is to substitute the $x$ from the second equation into the first equation. This is because in the second equation the $x$ is already isolated. To use substitution, we need a variable which is isolated (alone on one side of the equation).

In this case, the substitution looks like this:

Combining these equations by substitution, we get:

Since the second equation tells us that $x$ is equal to $x=- y - 3$, we can substitute it into the other equation in place of $x$ straight away.

The best choice is to substitute the $x$ from the second equation into the first equation because it is already isolated.

The best choice for substitution is based on which variable is easiest to isolate. In this case, you should look for the variable which is already isolated.

For this question, we are thinking about solving two equations 'simultaneously'. That means solving them together. One method for solving equations simultaneously is substitution. The substitution method works by using one of the equations to replace one of the variables in the other equation. We need to figure out which variable is the best choice to substitute for these two equations.

In these two equations, there are two $x$-terms and two $y$-terms:

The goal of substitution is to use one of the equations to remove a variable from the other equation.

In this case, the best choice is to substitute the $x$ from the second equation into the first equation. This is because in the second equation the $x$ is already isolated. To use substitution, we need a variable which is isolated (alone on one side of the equation).

In this case, the substitution looks like this:

Combining these equations by substitution, we get:

Since the second equation tells us that $x$ is equal to $x=- 4 y + 4$, we can substitute it into the other equation in place of $x$ straight away.

The best choice is to substitute the $x$ from the second equation into the first equation because it is already isolated.

Remember that the goal of elimination is to cancel either the $m$-terms or the $n$-terms. So you need to change one of the coefficients in the first equations so that you can cancel terms.

For this question we need to multiply the first equation by a number so that we can solve the equations by elimination. In the original equations, none of the coefficients were ready to cancel (because none of them are equal). The second equation was already multiplied by $-\text{4}$. We need to change one of the coefficients in the first equation so that we can cancel terms.

Specifically, we need to multiply $-\text{5}$ to get $-\text{16}$ so we can cancel the $m$-terms, or multiply the $-\text{4}$ to get $-\text{20}$ so we can cancel the $n$-terms.

The numbers in these equations point us toward cancelling the $n$-terms. This is because $-\text{20}$ is a multiple of $-\text{4}$. So if we multiply the first equation by $\text{5}$ the the coefficients will be equal.

The coefficients we want to cancel are equal. So eliminating the terms requires subtracting both sides of the equation:

The correct answer can be either of these numbers: $\text{5}$ or $-\text{5}$.

Remember that the goal of elimination is to cancel either the $x$-terms or the $y$-terms. So you need to change one of the coefficients in the second equations so that you can cancel terms.

For this question we need to multiply the second equation by a number so that we can solve the equations by elimination. In the original equations, none of the coefficients were ready to cancel (because none of them are equal). The first equation was already multiplied by $\text{4}$. We need to change one of the coefficients in the second equation so that we can cancel terms.

Specifically, we need to multiply $\text{4}$ to get $\text{20}$ so we can cancel the $x$-terms, or multiply the $\text{3}$ to get $\text{16}$ so we can cancel the $y$-terms.

The numbers in these equations point us toward cancelling the $x$-terms. This is because $\text{20}$ is a multiple of $\text{4}$. So if we multiply the second equation by $\text{5}$ the the coefficients will be equal.

The coefficients we want to cancel are equal. So eliminating the terms requires subtracting both sides of the equation:

The correct answer can be either of these numbers: $\text{5}$ or $-\text{5}$.

Remember that the goal of elimination is to cancel either the $k$-terms or the $p$-terms. So you need to change one of the coefficients in the first equations so that you can cancel terms.

For this question we need to multiply the first equation by a number so that we can solve the equations by elimination. In the original equations, none of the coefficients were ready to cancel (because none of them are equal). The second equation was already multiplied by $-\text{2}$. We need to change one of the coefficients in the first equation so that we can cancel terms.

Specifically, we need to multiply $\text{5}$ to get $-\text{6}$ so we can cancel the $k$-terms, or multiply the $-\text{2}$ to get $\text{6}$ so we can cancel the $p$-terms.

The numbers in these equations point us toward cancelling the $p$-terms. This is because $\text{6}$ is a multiple of $-\text{2}$. So if we multiply the first equation by $\text{3}$ the the coefficients will be equal and opposite.

The coefficients we want to cancel are equal and opposite. So eliminating the terms requires adding both sides of the equation:

The correct answer can be either of these numbers: $-\text{3}$ or $\text{3}$.

You need to find the coordinates of the point where the lines intersect.

This question is about two equations. We need to find the values of $x$ and $y$ which solve the equations 'simultaneously'. This means we want one pair of $x$ and $y$ values which solves both equations.

There are different methods to solve equations simultaneously. In this case we can use the graph: the solutions we want are the coordinates where the lines intersect. The intersection point is very special - it is the only point shared by both lines. And the coordinates of that point are the only numbers which solve both equations! In this case the lines intersect at the point $\left(1; -2\right)$.

We can prove that this is correct by substituting the values into each of the equations:

and

Perfect - the values $x = 1$ and $y = -2$ solve the equations simultaneously.

The correct answers are $x = 1$ and $y = -2$.

You need to find the coordinates of the point where the lines intersect.

This question is about two equations. We need to find the values of $x$ and $y$ which solve the equations 'simultaneously'. This means we want one pair of $x$ and $y$ values which solves both equations.

There are different methods to solve equations simultaneously. In this case we can use the graph: the solutions we want are the coordinates where the lines intersect. The intersection point is very special - it is the only point shared by both lines. And the coordinates of that point are the only numbers which solve both equations! In this case the lines intersect at the point $\left(3; 4\right)$.

We can prove that this is correct by substituting the values into each of the equations:

and

Perfect - the values $x = 3$ and $y = 4$ solve the equations simultaneously.

The correct answers are $x = 3$ and $y = 4$.

You need to find the coordinates of the point where the lines intersect.

This question is about two equations. We need to find the values of $x$ and $y$ which solve the equations 'simultaneously'. This means we want one pair of $x$ and $y$ values which solves both equations.

There are different methods to solve equations simultaneously. In this case we can use the graph: the solutions we want are the coordinates where the lines intersect. The intersection point is very special - it is the only point shared by both lines. And the coordinates of that point are the only numbers which solve both equations! In this case the lines intersect at the point $\left(-2; 1\right)$.

We can prove that this is correct by substituting the values into each of the equations:

and

Perfect - the values $x = -2$ and $y = 1$ solve the equations simultaneously.

The correct answers are $x = -2$ and $y = 1$.

The first thing you need to do (if you will use substitution to solve the pair of equations) is isolate one of the variables from one of the equations; then substitute what you get into the other equation.

When we solve simultaneous equations, we are working to find values for $y$ and $x$ which make both equations true!

We can solve these equations by substitution. Substitution requires isolating a variable. We need to look at the equations to decide which variable to isolate. We can see that the first equation looks good for this, because the $y$ has a coefficient of one:

Isolate $y$ in the first equation, which can be done in one step:

Now we can substitute the expression $3 x + 11$ into the second equation and solve:

Right now we have only half of the answer! To get the other half, we need to find the value of $y$. We can use the equation we got when we isolated $y$: it is the most convenient choice because $y$ is the subject of the equation.

The correct answers are $x = -5$ and $y = -4$.

The first thing you need to do (if you will use substitution to solve the pair of equations) is isolate one of the variables from one of the equations; then substitute what you get into the other equation.

When we solve simultaneous equations, we are working to find values for $x$ and $y$ which make both equations true!

We can solve these equations by substitution. Substitution requires isolating a variable. We need to look at the equations to decide which variable to isolate. We can see that the first equation looks good for this, because the $x$ has a coefficient of one:

Isolate $x$ in the first equation, which can be done in one step:

Now we can substitute the expression $4 y + 26$ into the second equation and solve:

Right now we have only half of the answer! To get the other half, we need to find the value of $x$. We can use the equation we got when we isolated $x$: it is the most convenient choice because $x$ is the subject of the equation.

The correct answers are $x = 10$ and $y = -4$.

The first thing you need to do (if you will use substitution to solve the pair of equations) is isolate one of the variables from one of the equations; then substitute what you get into the other equation.

When we solve simultaneous equations, we are working to find values for $y$ and $x$ which make both equations true!

We can solve these equations by substitution. Substitution requires isolating a variable. We need to look at the equations to decide which variable to isolate. We can see that the second equation looks good for this, because the $y$ has a coefficient of one:

Isolate $y$ in the second equation, which can be done in one step:

Now we can substitute the expression $2 x - 3$ into the first equation and solve:

Right now we have only half of the answer! To get the other half, we need to find the value of $y$. We can use the equation we got when we isolated $y$: it is the most convenient choice because $y$ is the subject of the equation.

The correct answers are $x = 4$ and $y = 5$.

You need to plot the graphs of the two equations on the same Cartesian plane.

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: $y = x^{2} + 6 x + 5$.

To plot this graph, we need to find the coordinates of the points where the graph will cross the $x$- and $y$-axes. These points are the $x$- and $y$-intercepts, respectively. We will then use these points to plot the graph.

The $y$-intercept is the quantity in the equation which is not multiplied by $x$. For this equation, the $y$-intercept is $\text{5}$. The coordinates of this point are $\left(0; 5\right)$.

The $x$-intercepts are the roots of the equation $x^{2} + 6 x + 5 = 0$. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{6}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{5}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{5}$. The two numbers are $\text{1}$ and $\text{5}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are $x = -1$ and $x = -5$. The coordinates of these roots are $\left(-1; 0 \right)$ and $\left(-5; 0 \right)$.

Using these $x$- and $y$-intercepts we will draw the graph:

As we did for the quadratic equation, we will first find the $y$-intercept. After that we will calculate root for the linear equation. We remember that the $y$-intercept is the quantity in the equation which is not multiplied by $x$.

The $y$-intercept is $-\text{10}$. Its coordinates are: $\left(0; -10\right)$.

We will work out the root of the equation $0 = - 2 x - 10$.

The root is: $x = -5$. It's coordinates are: $\left( -5; 0 \right)$.

We will now plot the equation $y = - 2 x - 10$ on the same Cartesian plane we have used for the graph of the quadratic equation.

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: $(-3; -4)$ and $(-5; 0)$.

We have used a graphical method to determine the solutions of the simultaneous equations: $y = x^{2} + 6 x + 5$ and $y = - 2 x - 10$. The solutions are the points where their graphs intersect.

The points of intersection are $(-3; -4)$ and $(-5; 0)$.

You need to plot the graphs of the two equations on the same Cartesian plane.

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: $y = x^{2} - 7 x + 6$.

To plot this graph, we need to find the coordinates of the points where the graph will cross the $x$- and $y$-axes. These points are the $x$- and $y$-intercepts, respectively. We will then use these points to plot the graph.

The $y$-intercept is the quantity in the equation which is not multiplied by $x$. For this equation, the $y$-intercept is $\text{6}$. The coordinates of this point are $\left(0; 6\right)$.

The $x$-intercepts are the roots of the equation $x^{2} - 7 x + 6 = 0$. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{7}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{6}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{6}$. The two numbers are $-\text{1}$ and $-\text{6}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are $x = 6$ and $x = 1$. The coordinates of these roots are $\left(6; 0 \right)$ and $\left(1; 0 \right)$.

Using these $x$- and $y$-intercepts we will draw the graph:

As we did for the quadratic equation, we will first find the $y$-intercept. After that we will calculate root for the linear equation. We remember that the $y$-intercept is the quantity in the equation which is not multiplied by $x$.

The $y$-intercept is $\text{2}$. Its coordinates are: $\left(0; 2\right)$.

We will work out the root of the equation $0 = - 2 x + 2$.

The root is: $x = 1$. It's coordinates are: $\left( 1; 0 \right)$.

We will now plot the equation $y = - 2 x + 2$ on the same Cartesian plane we have used for the graph of the quadratic equation.

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: $(4; -6)$ and $(1; 0)$.

We have used a graphical method to determine the solutions of the simultaneous equations: $y = x^{2} - 7 x + 6$ and $y = - 2 x + 2$. The solutions are the points where their graphs intersect.

The points of intersection are $(4; -6)$ and $(1; 0)$.

You need to plot the graphs of the two equations on the same Cartesian plane.

We must plot the graphs of the two equations on the same Cartesian plane. We will then need to find the points where the two graphs intersect. We will begin by plotting graph of the quadratic equation: $y = x^{2} + 3 x - 4$.

To plot this graph, we need to find the coordinates of the points where the graph will cross the $x$- and $y$-axes. These points are the $x$- and $y$-intercepts, respectively. We will then use these points to plot the graph.

The $y$-intercept is the quantity in the equation which is not multiplied by $x$. For this equation, the $y$-intercept is $\text{−4}$. The coordinates of this point are $\left(0; -4\right)$.

The $x$-intercepts are the roots of the equation $x^{2} + 3 x - 4 = 0$. We will calculate the roots of the equation through factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{3}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{4}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $-\text{4}$. The two numbers are $\text{4}$ and $-\text{1}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise. After that, we will factor the highest common factor.

If we equate each binomial to zero we get the roots of the quadratic equation. The roots are $x = 1$ and $x = -4$. The coordinates of these roots are $\left(1; 0 \right)$ and $\left(-4; 0 \right)$.

Using these $x$- and $y$-intercepts we will draw the graph:

As we did for the quadratic equation, we will first find the $y$-intercept. After that we will calculate root for the linear equation. We remember that the $y$-intercept is the quantity in the equation which is not multiplied by $x$.

The $y$-intercept is $\text{1}$. Its coordinates are: $\left(0; 1\right)$.

We will work out the root of the equation $0 = - x + 1$.

The root is: $x = 1$. It's coordinates are: $\left( 1; 0 \right)$.

We will now plot the equation $y = - x + 1$ on the same Cartesian plane we have used for the graph of the quadratic equation.

As we can see from the second graph, there are two points of intersection. We have marked each of these points with a black dot. The coordinates of these points are: $(1; 0)$ and $(-5; 6)$.

We have used a graphical method to determine the solutions of the simultaneous equations: $y = x^{2} + 3 x - 4$ and $y = - x + 1$. The solutions are the points where their graphs intersect.

The points of intersection are $(1; 0)$ and $(-5; 6)$.

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate $y$.

Now we need to substitute the entire right side of the equation, $(- x + 12)$, into the quadratic equation in place of $y$. Be sure to use brackets! Then simplify the equation to get standard form.

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two $x$-values.

Cool! However, there is still work to do. We must now find the values of $y$ for these equations. We must substitute in the values $x = 4$ and $x = 2$. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally we have the answers: if $x = 4$ then $y = 8$ and if $x = 2$ then $y = 10$.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: $\left(4 ; 8\right)$ and $\left(2 ; 10\right)$.

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate $x$.

Now we need to substitute the entire right side of the equation, $(3 y + 10)$, into the quadratic equation in place of $x$. Be sure to use brackets! Then simplify the equation to get standard form.

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two $y$-values.

Cool! However, there is still work to do. We must now find the values of $x$ for these equations. We must substitute in the values $y = 1$ and $y = 2$. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally we have the answers: if $x = 13$ then $y = 1$ and if $x = 16$ then $y = 2$.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: $\left(13 ; 1\right)$ and $\left(16 ; 2\right)$.

Begin by rearranging the linear equation to isolate one of the variables. Then you will be able to substitute the linear equation into the quadratic equation and solve from there.

The first step is to arrange the linear equation so that one of the variables is alone on one side of the equation. It does not matter which variable we use, so it is best to select the easiest choice. In this case, that means that we will isolate $x$.

Now we need to substitute the entire right side of the equation, $(- 6 y + 17)$, into the quadratic equation in place of $x$. Be sure to use brackets! Then simplify the equation to get standard form.

This equation can be solved by factorisation. (You can also solve the equation with the quadratic formula if you want.) This will give us two $y$-values.

Cool! However, there is still work to do. We must now find the values of $x$ for these equations. We must substitute in the values $y = -3$ and $y = 3$. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally we have the answers: if $x = 35$ then $y = -3$ and if $x = -1$ then $y = 3$.

The graph here shows the two equations in this problem, one linear and the other quadratic. The graphs intersect in two places, which are shown in red - these points of intersection are the solutions to the simultaneous equation system.

The solutions are: $\left(35 ; -3\right)$ and $\left(-1 ; 3\right)$.

You need to start by substituting $2 x - 4$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{8}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{12}$. This is the product of the coefficient of the first term $\text{2}$ and the value of the last term $\text{6}$. The two numbers are $-\text{2}$ and $-\text{6}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\text{3}$ and $\text{1}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = 3$:

For $x = 1$:

The solutions are $x = 3$ together with $y = 2$ , and $x = 1$ with $y = -2$. In coordinates pairs, the solutions are $\left(3; 2\right)$ and $\left(1; -2\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(3; 2\right)$ and $\left(1; -2\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(3; 2\right)$ and $\left(1; -2\right)$.

You need to start by substituting $x - 3$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{5}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{4}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{4}$. The two numbers are $-\text{1}$ and $-\text{4}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\text{4}$ and $\text{1}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = 4$:

For $x = 1$:

The solutions are $x = 4$ together with $y = 1$ , and $x = 1$ with $y = -2$. In coordinates pairs, the solutions are $\left(4; 1\right)$ and $\left(1; -2\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(4; 1\right)$ and $\left(1; -2\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(4; 1\right)$ and $\left(1; -2\right)$.

You need to start by substituting $3 x - 8$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{12}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{32}$. This is the product of the coefficient of the first term $\text{2}$ and the value of the last term $\text{16}$. The two numbers are $-\text{4}$ and $-\text{8}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\text{4}$ and $\text{2}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = 4$:

For $x = 2$:

The solutions are $x = 4$ together with $y = 4$ , and $x = 2$ with $y = -2$. In coordinates pairs, the solutions are $\left(4; 4\right)$ and $\left(2; -2\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(4; 4\right)$ and $\left(2; -2\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(4; 4\right)$ and $\left(2; -2\right)$.

You need to start by substituting $x + 3$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{4}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{12}$. This is the product of the coefficient of the first term $\text{3}$ and the value of the last term $-\text{4}$. The two numbers are $\text{6}$ and $-\text{2}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\dfrac{2}{3}$ and $-\text{2}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = \dfrac{2}{3}$:

For $x = -2$:

The solutions are $x = \dfrac{2}{3}$ together with $y = \dfrac{11}{3}$ , and $x = -2$ with $y = 1$. In coordinate pairs, the solutions are $\left(\dfrac{2}{3}; \dfrac{11}{3}\right)$ and $\left(-2; 1\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(\dfrac{2}{3}; \dfrac{11}{3}\right)$ and $\left(-2; 1\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(\dfrac{2}{3}; \dfrac{11}{3}\right)$ and $\left(-2; 1\right)$.

You need to start by substituting $- x - 1$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{3}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{4}$. This is the product of the coefficient of the first term $\text{2}$ and the value of the last term $-\text{2}$. The two numbers are $\text{4}$ and $-\text{1}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\dfrac{1}{2}$ and $-\text{2}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = \dfrac{1}{2}$:

For $x = -2$:

The solutions are $x = \dfrac{1}{2}$ together with $y = -\dfrac{3}{2}$ , and $x = -2$ with $y = 1$. In coordinate pairs, the solutions are $\left(\dfrac{1}{2}; -\dfrac{3}{2}\right)$ and $\left(-2; 1\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(\dfrac{1}{2}; -\dfrac{3}{2}\right)$ and $\left(-2; 1\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(\dfrac{1}{2}; -\dfrac{3}{2}\right)$ and $\left(-2; 1\right)$.

You need to start by substituting $- 2 x - 1$ in place of $y$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. Then we will simplify.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $\text{5}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{6}$. This is the product of the coefficient of the first term $\text{3}$ and the value of the last term $-\text{2}$. The two numbers are $\text{6}$ and $-\text{1}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\dfrac{1}{3}$ and $-\text{2}$.

We will substitute each $x$-value into the linear equation. This will give us the corresponding values of $y$.

For $x = \dfrac{1}{3}$:

For $x = -2$:

The solutions are $x = \dfrac{1}{3}$ together with $y = -\dfrac{5}{3}$ , and $x = -2$ with $y = 3$. In coordinate pairs, the solutions are $\left(\dfrac{1}{3}; -\dfrac{5}{3}\right)$ and $\left(-2; 3\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(\dfrac{1}{3}; -\dfrac{5}{3}\right)$ and $\left(-2; 3\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(\dfrac{1}{3}; -\dfrac{5}{3}\right)$ and $\left(-2; 3\right)$.

You need to start by substituting $y - 4$ in place of $x$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $x$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{6}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{8}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{8}$. The two numbers are $-\text{2}$ and $-\text{4}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$-values of our solutions. The $y$-values are 4 and 2.

We will substitute each $y$-value into the linear equation. This will give us the corresponding values of $x$.

For $y = 4$:

For $y = 2$:

The solutions are $x = 0$ together with $y = 4$ , and $x = -2$ with $y = 2$. In coordinates pairs, the solutions are $\left(0; 4\right)$ and $\left(-2; 2\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(0; 4\right)$ and $\left(-2; 2\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(0; 4\right)$ and $\left(-2; 2\right)$.

You need to start by substituting $y - 4$ in place of $x$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $x$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{6}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{8}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{8}$. The two numbers are $-\text{2}$ and $-\text{4}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$-values of our solutions. The $y$-values are 4 and 2.

We will substitute each $y$-value into the linear equation. This will give us the corresponding values of $x$.

For $y = 4$:

For $y = 2$:

The solutions are $x = 0$ together with $y = 4$ , and $x = -2$ with $y = 2$. In coordinates pairs, the solutions are $\left(0; 4\right)$ and $\left(-2; 2\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(0; 4\right)$ and $\left(-2; 2\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(0; 4\right)$ and $\left(-2; 2\right)$.

You need to start by substituting $y + 1$ in place of $x$ in the quadratic equation and then simplify.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $x$ from the linear equation into the quadratic equation. Then we will simplify.

Now that we have substituted the linear equation into the quadratic one, it's time to solve the new quadratic equation.

We will now solve the new quadratic equation by factorisation. In this case, we will use the grouping method. For this method, we need two numbers that add to give $-\text{2}$. This is the coefficient of the middle term. The product of the two numbers must be $-\text{3}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $-\text{3}$. The two numbers are $\text{1}$ and $-\text{3}$. We will re-write the middle term, forming two terms, using these two numbers.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$-values of our solutions. The $y$-values are 3 and -1.

We will substitute each $y$-value into the linear equation. This will give us the corresponding values of $x$.

For $y = 3$:

For $y = -1$:

The solutions are $x = 4$ together with $y = 3$ , and $x = 0$ with $y = -1$. In coordinates pairs, the solutions are $\left(4; 3\right)$ and $\left(0; -1\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(4; 3\right)$ and $\left(0; -1\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(4; 3\right)$ and $\left(0; -1\right)$.

Solve the equations simultaneously to find their intersection points.

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation $y$ is equal to the quantity $\frac{x}{2} - \frac{3}{2}$; and at the same time, the quadratic equation also contains the value $y$. Substitute like this:

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is $\text{2}$. So multiply both sides of the equation by $\text{2}$ in order to cancel the denominators. Then arrange the equation in standard form.

Remember that if the quadratic coefficient is negative, we can multiply the equation by $\text{−1}$. That will make it easier to factorise.

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two $x$-values.

Great... but it isn't over yet. We must now find the values of $y$ for these equations. Substitute in the values $x = -2$ and $x = 3$ into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally! The solutions are $\left(-2 ; - \frac{5}{2}\right)$ and $\left(3 ; 0\right)$.

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are $\left(-2 ; - \frac{5}{2}\right)$ and $\left(3 ; 0\right)$.

Solve the equations simultaneously to find their intersection points.

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation $y$ is equal to the quantity $- \frac{16 x}{3} + \frac{4}{3}$; and at the same time, the quadratic equation also contains the value $y$. Substitute like this:

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is $\text{3}$. So multiply both sides of the equation by $\text{3}$ in order to cancel the denominators. Then arrange the equation in standard form.

In this case, there is a common factor of $\text{−4}$ for all of the terms. Divide out this factor before factorising.

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two $x$-values.

Great... but it isn't over yet. We must now find the values of $y$ for these equations. Substitute in the values $x = 0$ and $x = 4$ into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally! The solutions are $\left(0 ; \frac{4}{3}\right)$ and $\left(4 ; -20\right)$.

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are $\left(0 ; \frac{4}{3}\right)$ and $\left(4 ; -20\right)$.

Solve the equations simultaneously to find their intersection points.

The intersection points of two graphs are the points where the two graphs have the same answers (the same coordinates). That means this problem is actually the same as simultaneous equations. We can solve the equations using substitution.

We can substitute immediately: we know that for the linear equation $y$ is equal to the quantity $\frac{x}{3} + \frac{7}{3}$; and at the same time, the quadratic equation also contains the value $y$. Substitute like this:

Now we want to arrange the equation in standard form. But before we do that, it will be helpful to remove those fractions! The LCD of the equation is $\text{3}$. So multiply both sides of the equation by $\text{3}$ in order to cancel the denominators. Then arrange the equation in standard form.

We could solve this equation with the quadratic formula. But it can be factorised, so we will take that approach. This will give us two $x$-values.

Great... but it isn't over yet. We must now find the values of $y$ for these equations. Substitute in the values $x = -3$ and $x = 3$ into one of the equations to do this. We can use either equation to do this, so we will choose the easier option: the linear equation.

Finally! The solutions are $\left(-3 ; \frac{4}{3}\right)$ and $\left(3 ; \frac{10}{3}\right)$.

The graph here shows the two equations in this problem, and the points of intersection shown with red dots. You can see that the answers agree with the positions of the dots on the graph.

The solutions are $\left(-3 ; \frac{4}{3}\right)$ and $\left(3 ; \frac{10}{3}\right)$.

You need to substitute $y - 1$ in place of $x$ in the quadratic equation $y = x^{2} + 2 x - 5$.

We have two equations that we will solve by substitution. The equations are:

We need to substitute $y - 1$ from the linear equation in place of $x$ in the quadratic equation.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for $y$. The question does not ask us to solve the simultaneous equations.

The correct substitution step is $y=(y - 1)^2+2(y - 1)-5$ which is choice D.

You need to substitute $- 3 x - 4$ in place of $y$ in the quadratic equation $y = x^{2} - 5 x - 7$.

We have two equations that we will solve by substitution. The equations are:

We need to substitute $- 3 x - 4$ from the linear equation in place of $y$ in the quadratic equation.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for $x$. The question does not ask us to solve the simultaneous equations.

The correct substitution step is $- 3 x - 4=x^{2} - 5 x - 7$ which is choice C.

You need to substitute $- 3 x - 10$ in place of $y$ in the quadratic equation $y = x^{2} - 8 x - 6$.

We have two equations that we will solve by substitution. The equations are:

We need to substitute $- 3 x - 10$ from the linear equation in place of $y$ in the quadratic equation.

That's it! That is the first step we need to take. Once this is done, we can proceed to solve for $x$. The question does not ask us to solve the simultaneous equations.

The correct substitution step is $- 3 x - 10=x^{2} - 8 x - 6$ which is choice B.

You need to start by substituting $- 2 x - 4$ in place of $y$ in the quadratic equation and simplify. You must expect a special kind of solution.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is $y = x^{2} + 4 x + 5$ and the linear equation is $y = - 2 x - 4$. This means we expect two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We will factorise using the grouping method. For this method, we need two numbers that add to give $\text{6}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{9}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{9}$. If we take $\text{3}$ which is half of the value of $\text{6}$ (the coefficient of the middle term) and square it, we get the product $\text{9}$. In this case we will use only one number to factorise, that is, $\text{3}$. We will re-write the middle term, forming two terms, using this value.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, $-\text{3}$. This root is the $x$-value of our solution. This $x$-value will have one corresponding $y$ value.

Now that we have the value of $x$, we will substitute it into the linear equation to calculate the corresponding value of $y$.

The solution is $x = -3$ together with $y = 2$. Using coordinate pairs, the solution is $\left(-3; 2\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-3; 2\right)$. This is the same solution we have calculated. This means we are $\text{100}\%$ correct.

The solution is $\left(-3; 2\right)$.

You need to start by substituting $- 3 x - 1$ in place of $y$ in the quadratic equation and simplify. You must expect a special kind of solution.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is $y = 3 x^{2} - x$ and the linear equation is $y = - 3 x - 1$. This means we expect two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We need two numbers that add to give $\text{2}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{3}$. This is the product of the coefficient of the first term $\text{3}$ and the value of the last term $\text{1}$. In this case, there are no such numbers. We will now use the quadratic formula to solve this equation.

The solution for a quadratic equation of the form $ax^2 + bx + c = 0$:

Before we can use the quadratic formula, we need to identify the coefficients $a$, $b$ and $c$ from the equation $3 x^{2}+2 x+1 = 0$ These are:

We will substitute these values into the formula and simplify.

The number under the square root sign is negative. This means that our quadratic equation has no real solution. This means that $y = 3 x^{2} - x$ and $y = - 3 x - 1$ cannot be solved simultaneously.

We can confirm our result by plotting the graph of each equation on the same set of axes. We do not expect these graphs to intersect.

Aha! The two graphs do not have an intersection point. This means that indeed the two equations have no solution.

The final answer is no solution.

You need to start by substituting $- 3 x + 1$ in place of $y$ in the quadratic equation and simplify. You must expect a special kind of solution.

We must solve the two equations simultaneously. In this case, we have a quadratic equation and a linear equation. The quadratic equation is $y = x^{2} - x + 2$ and the linear equation is $y = - 3 x + 1$. This means we expect two pairs of $x$ and $y$ values which solve both equations.

We will substitute the value of $y$ from the linear equation into the quadratic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

We will factorise using the grouping method. For this method, we need two numbers that add to give $\text{2}$. This is the coefficient of the middle term. The product of the two numbers must be $\text{1}$. This is the product of the coefficient of the first term $\text{1}$ and the value of the last term $\text{1}$. If we take $\text{1}$ which is half of the value of $\text{2}$ (the coefficient of the middle term) and square it, we get the product $\text{1}$. In this case we will use only one number to factorise, that is, $\text{1}$. We will re-write the middle term, forming two terms, using this value.

We now group the first two terms together and group the last terms together and factorise.

The last thing now is to factor the highest common factor.

That's it! If we equate each binomial to zero we get the roots of the quadratic equation. In this case we have one root, $-\text{1}$. This root is the $x$-value of our solution. This $x$-value will have one corresponding $y$ value.

Now that we have the value of $x$, we will substitute it into the linear equation to calculate the corresponding value of $y$.

The solution is $x = -1$ together with $y = 4$. Using coordinate pairs, the solution is $\left(-1; 4\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-1; 4\right)$. This is the same solution we have calculated. This means we are $\text{100}\%$ correct.

The solution is $\left(-1; 4\right)$.

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.

We need to solve the following pair of equations for $x$ and $y$:

First, we can substitute $- 5 y$ in place of $x$ in the second equation:

We can simplify this equation to solve for $y$:

$y$ can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about $y^{2}$. So we have to consider both the positive and the negative values as solutions to the equations.

Each of the $y$-values that we have found will have a corresponding value of $x$, which can be found from the equation $x = - 5 y$.

For $y = 2$:

And for $y = -2$:

So the $x$-values are $x = -10 \text{ or } x = 10$.

Therefore the correct coordinate pairs are $(-10; 2)$ and $(10; -2)$.

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.

We need to solve the following pair of equations for $x$ and $y$:

First, we can substitute $5 y$ in place of $x$ in the second equation:

We can simplify this equation to solve for $y$:

$y$ can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about $y^{2}$. So we have to consider both the positive and the negative values as solutions to the equations.

Each of the $y$-values that we have found will have a corresponding value of $x$, which can be found from the equation $x = 5 y$.

For $y = 3$:

And for $y = -3$:

So the $x$-values are $x = 15 \text{ or } x = -15$.

Therefore the correct coordinate pairs are $(15; 3)$ and $(-15; -3)$.

You may find it helpful to revise simultaneous equations in the Everything Maths textbook.

We need to solve the following pair of equations for $x$ and $y$:

First, we can substitute $2 y$ in place of $x$ in the second equation:

We can simplify this equation to solve for $y$:

$y$ can have either a positive or a negative value. We cannot know for sure which one to pick, since we only have information about $y^{2}$. So we have to consider both the positive and the negative values as solutions to the equations.

Each of the $y$-values that we have found will have a corresponding value of $x$, which can be found from the equation $x = 2 y$.

For $y = 2$:

And for $y = -2$:

So the $x$-values are $x = 4 \text{ or } x = -4$.

Therefore the correct coordinate pairs are $(4; 2)$ and $(-4; -2)$.

You need to start by substituting $- x - 2$ in place of $y$ in the hyperbolic equation and then solve for $x$.

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $y$ from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $-\text{1}$ and $-\text{9}$.

By using the values of $x$, we will calculate the corresponding values of $y$. We will do so by substituting each $x$-value into the linear equation.

For $x = -1$:

For $x = -9$:

The solutions are $x = -1$ together with $y = -1$ , and $x = -9$ with $y = 7$. Using coordinate pairs, the solutions are $\left(-1; -1\right)$ and $\left(-9; 7\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(-1; -1\right)$ and $\left(-9; 7\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(-1; -1\right)$ and $\left(-9; 7\right)$.

You need to start by substituting $y - 1$ in place of $x$ in the hyperbolic equation and then solve for $y$.

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $x$ from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $y$-values of our solutions. The $y$-values are $\text{6}$ and $\text{1}$.

By using the values of $y$, we will calculate the corresponding values of $x$. We will do so by substituting each $y$-value into the linear equation.

For $y = 6$:

For $y = 1$:

The solutions are $x = 5$ together with $y = 6$ , and $x = 0$ with $y = 1$. Using coordinate pairs, the solutions are $\left(5; 6\right)$ and $\left(0; 1\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(5; 6\right)$ and $\left(0; 1\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(5; 6\right)$ and $\left(0; 1\right)$.

You need to start by substituting $- 2 x + 7$ in place of $y$ in the hyperbolic equation and then solve for $x$.

We must solve the two equations simultaneously. In this case, we have a hyperbolic equation and a linear equation. The equations are:

This means we want two pairs of $x$ and $y$-values which solve both equations.

We will substitute the value of $y$ from the linear equation into the hyperbolic equation. After that, we will then simplify the new equation.

We now need to solve the new quadratic equation. We will solve this quadratic equation by factorisation.

If we equate each binomial to zero we get the roots of the quadratic equation. These roots are the $x$-values of our solutions. The $x$-values are $\text{6}$ and $-\text{1}$.

By using the values of $x$, we will calculate the corresponding values of $y$. We will do so by substituting each $x$-value into the linear equation.

For $x = 6$:

For $x = -1$:

The solutions are $x = 6$ together with $y = -5$ , and $x = -1$ with $y = 9$. Using coordinate pairs, the solutions are $\left(6; -5\right)$ and $\left(-1; 9\right)$.

We can plot the two equations on the same graph to check if our solutions are correct. The points where the two graphs intersect are the solutions of the equations.

The two graphs intersect at the points $\left(6; -5\right)$ and $\left(-1; 9\right)$. These are the same solutions we have calculated. This means we are $\text{100}\%$ correct.

The solutions are $\left(6; -5\right)$ and $\left(-1; 9\right)$.

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let $n_{1}$ represent one of the numbers and $n_{2}$ represent the other number". So we can break up the first fact like this:

The correct equation for the first fact is $n_{1} \times n_{2} = 42$.

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like $\text{10}$ and $\text{11}$, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is $\text{1}$ more than the small number." Another is "the small number is $\text{1}$ less than the larger number." Let's use the second option.

This makes sense: consecutive numbers are separated by $\text{1}$. The equation shows that with the $-\text{1}$.

The correct answers are:

For the first equation, the key word is sum. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let $n_{1}$ represent one of the numbers and $n_{2}$ represent the other number". So we can break up the first fact like this:

The correct equation for the first fact is $n_{1} + n_{2} = 19$.

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like $\text{10}$ and $\text{11}$, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is $\text{1}$ more than the small number." Another is "the small number is $\text{1}$ less than the larger number." Let's use the first option.

This makes sense: consecutive numbers are separated by $\text{1}$. The equation shows that with the $\text{+1}$.

The correct answers are:

For the first equation, the key word is product. For the second equation the key word is consecutive. Use these key words to figure out what operations should be in each equation.

In this question we need to translate words into equations. This can be challenging. One useful approach is to look for important words which tell us what numbers and calculations to use. Here are some common words and what they mean when we write mathematical expressions and equations:

With these key words in mind, let's identify the key parts/words in each of these facts. Then we can translate each of the parts into maths.

The question says "let $n_{1}$ represent one of the numbers and $n_{2}$ represent the other number". So we can break up the first fact like this:

The correct equation for the first fact is $n_{1} \times n_{2} = 42$.

We can approach the second fact in a similar way. But the second fact does not include the word "is". It would be good if it includes "is" because that tells us where the equal sign belongs. So let's rewrite the statement to include the word "is".

The second fact says "the numbers are consecutive numbers". (Consecutive means that the numbers follow each other, like $\text{10}$ and $\text{11}$, or like the letters b and c in the alphabet.) There are different ways to rewrite this. One way is "the larger number is $\text{1}$ more than the small number." Another is "the small number is $\text{1}$ less than the larger number." Let's use the first option.

This makes sense: consecutive numbers are separated by $\text{1}$. The equation shows that with the $\text{+1}$.

The correct answers are:

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two helicopters must be equal to the total distance when the helicopters meet:

This question is about distances, speeds, and times. The equation connecting these values is

We want to know the amount of time needed for the helicopters to meet. Let the time taken be $t$. Then we can write an expression for the distance each of the helicopters travels:

For helicopter 1:

For helicopter 2:

Now we have three different equations. We can combine them using substitution to solve for the value of $t$.

The helicopters will meet after $\text{8}$ hours.

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two ships must be equal to the total distance when the ships meet:

This question is about distances, speeds, and times. The equation connecting these values is

We want to know the amount of time needed for the ships to meet. Let the time taken be $t$. Then we can write an expression for the distance each of the ships travels:

For ship 1:

For ship 2:

Now we have three different equations. We can combine them using substitution to solve for the value of $t$.

The ships will meet after $\text{1}$ hours.

Draw a simple picture to organise the information. Then work out equations to describe the information in the question.

Start by making a quick sketch of the situation - label everything you know!

Notice that the sum of the distances for the two cars must be equal to the total distance when the cars meet:

This question is about distances, speeds, and times. The equation connecting these values is

We want to know the amount of time needed for the cars to meet. Let the time taken be $t$. Then we can write an expression for the distance each of the cars travels:

For car 1:

For car 2:

Now we have three different equations. We can combine them using substitution to solve for the value of $t$.

The cars will meet after $\text{8}$ hours.

You need to choose variables to represent the things you want to find and write equations based on the information in the question.

The two things we want to know in this question are the prices for each of the items (a pizza and a hamburger). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the $\text{7}$ pizzas and $\text{6}$ hamburgers is $\text{N}\llap{=}\,\text{405}$." We can use the expression $7 p$ to represent the price of the $\text{7}$ pizzas. Similarly, the expression $6 h$ represents the price of the $\text{6}$ hamburgers. With these values we can write a full equation for the prices:

Now we can use the second point. It says, "a pizza costs $\text{N}\llap{=}\,\text{4}$ more than a hamburger." We can write this as an equation like this:

Now that we have two equations, we need to solve them simultaneously.

We could use elimination, but substitution is a better choice (because $p$ is already the subject). Substitute the second equation into the first equation and solve!

This means that the price of one hamburger is $\text{N}\llap{=}\,\text{29}$.

Finally, use the value we found for $h$ to find the value of $p$ (the price of a pizza). We can use either equation to do this, but the second one is easier to use (because $p$ is already isolated).

The price for one pizza is $\text{N}\llap{=}\,\text{33}$.

It is important to write the answer to a word problem as a complete sentence.

The price of the pizza is $\text{N}\llap{=}\,\text{33}$ while a hamburger costs $\text{N}\llap{=}\,\text{29}$.

You need to choose variables to represent the things you want to find and write equations based on the information in the question.

The two things we want to know in this question are the prices for each of the items (a salad and a milkshake). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the $\text{7}$ salads and $\text{8}$ milkshakes is $\text{N}\llap{=}\,\text{417}$." We can use the expression $7 s$ to represent the price of the $\text{7}$ salads. Similarly, the expression $8 m$ represents the price of the $\text{8}$ milkshakes. With these values we can write a full equation for the prices:

Now we can use the second point. It says, "a salad costs $\text{N}\llap{=}\,\text{6}$ more than a milkshake." We can write this as an equation like this:

Now that we have two equations, we need to solve them simultaneously.

We could use elimination, but substitution is a better choice (because $s$ is already the subject). Substitute the second equation into the first equation and solve!

This means that the price of one milkshake is $\text{N}\llap{=}\,\text{25}$.

Finally, use the value we found for $m$ to find the value of $s$ (the price of a salad). We can use either equation to do this, but the second one is easier to use (because $s$ is already isolated).

The price for one salad is $\text{N}\llap{=}\,\text{31}$.

It is important to write the answer to a word problem as a complete sentence.

The price of the salad is $\text{N}\llap{=}\,\text{31}$ while a milkshake costs $\text{N}\llap{=}\,\text{25}$.

You need to choose variables to represent the things you want to find and write equations based on the information in the question.

The two things we want to know in this question are the prices for each of the items (a sandwich and a hamburger). To begin, we can pick a variable for each of these numbers. It is helpful to pick variables which remind you about the things in the question:

Next we need to write equations based on what the question tells us. In other words, we need to translate the words in the question into equations.

The first point says that "the total cost for the $\text{4}$ sandwiches and $\text{3}$ hamburgers is $\text{N}\llap{=}\,\text{221}$." We can use the expression $4 s$ to represent the price of the $\text{4}$ sandwiches. Similarly, the expression $3 h$ represents the price of the $\text{3}$ hamburgers. With these values we can write a full equation for the prices:

Now we can use the second point. It says, "a sandwich costs $\text{N}\llap{=}\,\text{8}$ more than a hamburger." We can write this as an equation like this:

Now that we have two equations, we need to solve them simultaneously.

We could use elimination, but substitution is a better choice (because $s$ is already the subject). Substitute the second equation into the first equation and solve!

This means that the price of one hamburger is $\text{N}\llap{=}\,\text{27}$.

Finally, use the value we found for $h$ to find the value of $s$ (the price of a sandwich). We can use either equation to do this, but the second one is easier to use (because $s$ is already isolated).

The price for one sandwich is $\text{N}\llap{=}\,\text{35}$.

It is important to write the answer to a word problem as a complete sentence.

The price of the sandwich is $\text{N}\llap{=}\,\text{35}$ while a hamburger costs $\text{N}\llap{=}\,\text{27}$.

You need to think about the meaning of the number you want. The question asks us for the price of each mango. Can the price be a decimal, like $\text{4.5}$? Can it be negative, like $-\text{5}$?

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be $\text{N}\llap{=}\,\text{6.75}$. Or $\text{N}\llap{=}\,\text{5}$. Or $\text{N}\llap{=}\,\text{19.87}$. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like $\text{N}\llap{=}\,\text{4.30}$, or must a price always be a whole number like $\text{N}\llap{=}\,\text{4.00}$? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs -$\text{N}\llap{=}\,\text{5.50}$. That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

1. Yes, it can be a decimal
2. No, it cannot be negative

You need to think about the meaning of the number you want. The question asks us for the price of each mango. Can the price be a decimal, like $\text{4.5}$? Can it be negative, like $-\text{5}$?

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be $\text{N}\llap{=}\,\text{6.75}$. Or $\text{N}\llap{=}\,\text{5}$. Or $\text{N}\llap{=}\,\text{19.87}$. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the price can be a decimal number or not. In other words, can the price of something be a decimal number like $\text{N}\llap{=}\,\text{4.30}$, or must a price always be a whole number like $\text{N}\llap{=}\,\text{4.00}$? Of course we know that many prices are not whole numbers! This is similar to the example above about airtime, which can be a decimal number. There is no reason why the price of an item must be a whole number.

The answer for the first question is: Yes, the price can be a decimal number.

The second question is similar, but this time we need to decide if the price can be negative, or must be positive. Prices cannot be negative. Imagine something which costs -$\text{N}\llap{=}\,\text{5.50}$. That is impossible!

The answer to the second question is: No, the price cannot be a negative number.

The correct answer choices are:

1. Yes, it can be a decimal
2. No, it cannot be negative

You need to think about the meaning of the number you want. The question asks us for the number of bananas Ben buys. Can the number of items be a decimal, like $\text{4.5}$? Can it be negative, like $-\text{5}$?

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, think about the airtime balance on a phone. It might be $\text{N}\llap{=}\,\text{6.75}$. Or $\text{N}\llap{=}\,\text{5}$. Or $\text{N}\llap{=}\,\text{19.87}$. The amount of airtime can be a decimal value - it does not have to be a whole number.

For the first question we must decide if the number of items can be a decimal number or not. In other words, can the number of bananas be a decimal number like $\text{3.7}$ bananas? Or must it be a whole number like $\text{4}$ bananas? This is different from the airtime example above: airtime can be a decimal number, but it is not possible to buy $\text{3.5}$ bananas! (You might think, "Wait! I can buy $\text{3.5}$ bananas, just cut one of them in half." But no one should expect to do this while shopping.)

The answer for the first question is: No, the number of items cannot be a decimal number.

The second question is similar, but this time we need to decide if the number of items can be negative, or must be postive. The number of bananas cannot be negative. If Ben did not buy any bananas at all, then the number of bananas is zero. But is it impossible for him to buy $-\text{3}$ bananas.

The answer to the second question is: No, the number of items cannot be a negative number.

The correct answer choices are:

1. No, it must be an integer
2. No, it cannot be negative

You need to think about the meaning of the number you want. The question asks us for the width of the rectangle. Can the width of the rectangle be a decimal, like $\text{4.5}$? Can it be negative, like $-\text{5}$?

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be $-\text{150}$ days of school, shame. Similarly, it is not possible to have $\text{176.3562}$ days of school. What type of numbers are realistic for the width of the rectangle?

For the first question we must decide if the width of the rectangle can be a decimal number. Is that fine or is it impossible?

In other words, can a distance be a decimal number like $\text{3.5}\ \text{cm}$, or must a distance be a whole number like $\text{4}\ \text{cm}$? The answer is that there is no reason a distance must be limited to whole number values. This is different from the number of days in the school year. It is not possible to have $\text{3.5}$ days of school, but a rectangle certainly can have a width of $\text{3.5}\ \text{cm}$. Or $\text{5.927}\ \text{cm}$. There is no reason why the width of the rectangle must be a whole number.

The answer for the first question is: Yes, it can be a decimal, the width of the rectangle can be a decimal number.

The second question is similar, but this time we need to decide if width of the rectangle can be negative, or must be positive. This is straightfoward: distances cannot be negative. Imagine a room $-\text{5}$ metres wide. No way!

The answer for the second question is: No, the width of the rectangle cannot be a negative number.

The correct answer choices are:

1. Yes, it can be a decimal
2. No, it cannot be negative

You need to think about the meaning of the number you want. The question asks us for the width of the rectangle. Can the width of the rectangle be a decimal, like $\text{4.5}$? Can it be negative, like $-\text{5}$?

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be $-\text{150}$ days of school, shame. Similarly, it is not possible to have $\text{176.3562}$ days of school. What type of numbers are realistic for the width of the rectangle?

For the first question we must decide if the width of the rectangle can be a decimal number. Is that fine or is it impossible?

In other words, can a distance be a decimal number like $\text{3.5}\ \text{cm}$, or must a distance be a whole number like $\text{4}\ \text{cm}$? The answer is that there is no reason a distance must be limited to whole number values. This is different from the number of days in the school year. It is not possible to have $\text{3.5}$ days of school, but a rectangle certainly can have a width of $\text{3.5}\ \text{cm}$. Or $\text{5.927}\ \text{cm}$. There is no reason why the width of the rectangle must be a whole number.

The answer for the first question is: Yes, it can be a decimal, the width of the rectangle can be a decimal number.

The second question is similar, but this time we need to decide if width of the rectangle can be negative, or must be positive. This is straightfoward: distances cannot be negative. Imagine a room $-\text{5}$ metres wide. No way!

The answer for the second question is: No, the width of the rectangle cannot be a negative number.

The correct answer choices are:

1. Yes, it can be a decimal
2. No, it cannot be negative

You need to think about the meaning of the number you want. The question asks us for the number of rectangles in the first figure of Danjuma's diagram. Can the number of rectangles be a decimal, like $\text{4.5}$? Can it be negative, like $-\text{5}$?

This question does not ask us to solve the word problem. Instead we need to think about what kind of answer is possible. For example, if we want to find the number of days in the school year, the answer must be a positive integer. There will never be $-\text{150}$ days of school, shame. Similarly, it is not possible to have $\text{176.3562}$ days of school. What type of numbers are realistic for the number of rectangles?

For the first question we must decide if the number of rectangles can be a decimal number. Is that fine or is it impossible?

In other words, can the number of rectangles be a decimal number like $\text{3.7}$, or must the number be a whole number like $\text{4}$? Just like the number of days in the school year, the number of rectangles cannot be a decimal value. $\text{3.7}$ rectangles does not make sense! The number of rectangles must be a whole number! (You might think, "That's wrong, we can just draw part of a rectangle to get $\text{3.7}$ rectangles." But if the shape is not complete, it is not a rectangle.)

The answer for the first question is: No, it must be an integer, the number of rectangles cannot be a decimal number.

The second question is similar, but this time we need to decide if number of rectangles can be negative, or must be positive. The number of rectangles cannot be negative. If Danjuma drew any rectangles at all, then the answer to this question must be positive. There is no such thing as $-\text{4}$ rectangles.

The answer for the second question is: No, the number of rectangles cannot be a negative number.

The correct answer choices are:

1. No, it must be an integer
2. No, it cannot be negative

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

This means that Atinuke's mark is $\text{75}$.

Now we can use Atinuke's mark and one of the equations we have to find Fezekile's mark. The easiest way to do that is to substitute Atinuke's mark back into the first equation:

The students achieved these marks: Fezekile earned $\text{87}$ and Atinuke earned $\text{75}$.

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

This means that Danjuma's mark is $\text{77}$.

Now we can use Danjuma's mark and one of the equations we have to find Kamogelo's mark. The easiest way to do that is to substitute Danjuma's mark back into the first equation:

The students achieved these marks: Kamogelo earned $\text{84}$ and Danjuma earned $\text{77}$.

Read through the question carefully and underline the important information. Then you will need to pick variables to represent each of the unknown facts from the question: the two marks. Use these variables to write down two equations which summarize the information in the question.

We need to figure out the marks the students got on their tests. The first thing to do is to pick variables for each student's mark. It is helpful to pick variables which match the information we want, for example:

Now we can write equations with those variables. There are two pieces of information we have about the marks, which lead to two equations:

We can solve these equations simultaneously. Substitute the first equation into the second equation and solve. (You can solve these equations using elimination if you prefer.)

This means that Amarachi's mark is $\text{62}$.

Now we can use Amarachi's mark and one of the equations we have to find Kungawo's mark. The easiest way to do that is to substitute Amarachi's mark back into the first equation:

The students achieved these marks: Kungawo earned $\text{70}$ and Amarachi earned $\text{62}$.

This question is about two unknown numbers. And we need to find one of them (the larger one). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a sum of $\text{14}$. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

The first fact given in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, $\text{10}$ and $\text{12}$ are consecutive even numbers. Since we defined $n_{1}$ as the smaller number, $n_{2}$ must be $\text{2}$ more than $n_{1}$.

The "$\text{+2}$" skips the odd integer which sits between $n_{1}$ and $n_{2}$.

The second fact about the numbers tells us that "the sum of the numbers is $\text{14}$". Remember that sum means addition. So:

We can now use these two equations to find the answer to the question. But remember that we only need to find the larger number (we do not need both of them). That means we need to find the value of $n_{2}$.

We can solve this using substitution. However, remember that we want the value of $n_2$ (the larger number). We can start by rearranging the first equation to make $n_1$ the subject. Then the substitution step will remove $n_1$ from the second equation and we can solve for $n_2$. (This is not required - it just makes the solution faster.)

Now substitute $n_2 - 2$ into the other equation and solve for $n_2$.

The result is $n_{2} = 8$. Notice that this means that the other number, $n_{1}$, must be $\text{6}$, because $n_{2} = n_{1} + 2$. This is perfect, because we know that the sum of the numbers is $\text{14}$, and $8 + 6 = 14$.

The larger number is $\text{8}$.

This question is about two unknown numbers. And we need to find one of them (the larger one). We know certain things about these numbers: they are positive, they are consecutive even integers, and they have a sum of $\text{18}$. We can solve this question using simultaneous equations.

The first thing to do is pick variables to represent the two numbers. Then we can write equations using those variables. It is usually helpful to pick variables which represent things we want to find. In this case, we are looking for two numbers, so these are good choices:

The first fact given in the question says that the numbers are "consecutive even integers". So both of the numbers are even, and they come one after another. For example, $\text{10}$ and $\text{12}$ are consecutive even numbers. Since we defined $n_{1}$ as the smaller number, $n_{2}$ must be $\text{2}$ more than $n_{1}$.

The "$\text{+2}$" skips the odd integer which sits between $n_{1}$ and $n_{2}$.

The second fact about the numbers tells us that "the sum of the numbers is $\text{18}$". Remember that sum means addition. So:

We can now use these two equations to find the answer to the question. But remember that we only need to find the larger number (we do not need both of them). That means we need to find the value of $n_{2}$.

We can solve this using substitution. However, remember that we want the value of $n_2$ (the larger number). We can start by rearranging the first equation to make $n_1$ the subject. Then the substitution step will remove $n_1$ from the second equation and we can solve for $n_2$. (This is not required - it just makes the solution faster.)